# Differantiation proof question exlanation problem

1. Jan 21, 2009

### transgalactic

if f(x) is differentiable on x_0
prove that

??

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

now let $u = x+h \Rightarrow x = u-h$

So $f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h}$

Since $h \to 0, f'(u-h) \to f'(u)$

Therefore $f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}$

$f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)$

$2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h}$

Hence, $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}$

i understand this solution but
how to pick those variables??
and why it didnt use the point x_0
that they presented

2. Jan 21, 2009

### transgalactic

there is another definition to limit that uses x_0 that i was given
$f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

is there a way to use this definition like you did before
what variables to pick so it will go on the "h" defnition??